# How do you solve the quadratic using the quadratic formula given 3x^2-11x=8-14x over the set of complex numbers?

Aug 30, 2016

Solution is $- \frac{1}{2} - \frac{\sqrt{105}}{6}$ and $- \frac{1}{2} + \frac{\sqrt{105}}{6}$

#### Explanation:

For the equation $a {x}^{2} + b x + c = 0$, the roots are given by $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$. As can be seen type of roots depend upon discriminant ${b}^{2} - 4 a c$. If it is negative, roots are complex and if it is a complete square (and coefficients are rational), roots are rational.

In the equation $3 {x}^{2} - 11 x = 8 - 14 x$ or $3 {x}^{2} + 3 x - 8 = 0$, the discriminant is (3)^2-4×3×(-8)=9+96=105, which is not a complete square and hence roots are irrational (not complex as mentioned in question).

Roots are x=(-3+-sqrt105)/(2×3)

= $\frac{- 3 \pm \sqrt{105}}{6}$

= $- \frac{1}{2} \pm \frac{\sqrt{105}}{6}$

Hence solution is $- \frac{1}{2} - \frac{\sqrt{105}}{6}$ and $- \frac{1}{2} + \frac{\sqrt{105}}{6}$.